$A=\left[\begin{array}{rr}-34 & -8 & 8 & -13 & 6 \\1 & 32 & 19 & 4 & -7 \\-8 &-12 &24 & -35 & -18 \\6 &18 &-11 & 1 & 8\end{array}\right]$ $A_{4,1}=$
Explanation: Background An $m\times n$ matrix has $m$ rows and $n$ columns. $A=\left[\begin{array}{rr}A_{1,1} & \cdots & A_{1,n} \\\\\vdots \ & \ddots & \vdots \\\\A_{m,1} &\cdots &A_{m,n}\end{array}\right]$ Therefore, the entry $A_{{c},{d}}$ is located on row ${c}$ and column ${d}$. Finding $A_{4,1}$ $A_{{4},{1}}$ is located on row ${4}$ of $A$ : $\left[\begin{array}{rr}-34 & -8 & 8 & -13 & 6 \\1 & 32 & 19 & 4 & -7 \\-8 &-12 &24 & -35 & -18 \\ {6} & {18} & {-11} & {1} & {8}\end{array}\right]$ $A_{{4},{1}}$ is also located on column ${1}$ of $A$. $\left[\begin{array}{rr}{-34} & -8 & 8 & -13 & 6 \\{1} & 32 & 19 & 4 & -7 \\{-8} &-12 &24 & -35 & -18 \\ {\text{6}} & {18} & {-11} & {1} & {8}\end{array}\right]$ Therefore, $A_{{4},{1}}={6}$. Summary $A_{4,1}=6$